Are these numbers sufficient for calculating $P(alpha|beta,gamma)$?

1

$begingroup$

Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?

My solution:

$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$

Since $beta$ and $gamma$ are independent given $alpha$, we can get:

$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$

Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.

Am I right?

probability conditional-probability

share|cite|improve this question

asked Jan 17 at 14:35

lernerlerner

314217

$endgroup$

add a comment | 

1

$begingroup$

Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?

My solution:

$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$

Since $beta$ and $gamma$ are independent given $alpha$, we can get:

$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$

Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.

Am I right?

probability conditional-probability

share|cite|improve this question

asked Jan 17 at 14:35

lernerlerner

314217

$endgroup$

add a comment | 

1

1

1

$begingroup$

Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?

My solution:

$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$

Since $beta$ and $gamma$ are independent given $alpha$, we can get:

$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$

Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.

Am I right?

probability conditional-probability

share|cite|improve this question

asked Jan 17 at 14:35

lernerlerner

314217

$endgroup$

Given three probabilities $Pr(beta|alpha)$, $Pr(gamma|alpha)$ and $Pr(alpha)$; and that $beta$ and $gamma$ are conditionally independent given $alpha$, can $Pr(alpha|beta, gamma)$ be calculated?

My solution:

$$begin{align}
Pr(alpha|beta, gamma) &= frac{Pr(alpha, beta, gamma)}{Pr(beta, gamma)} \
&= frac{Pr(alpha) Pr(beta, gamma|alpha)}{Pr(beta, gamma)}
end{align}$$

Since $beta$ and $gamma$ are independent given $alpha$, we can get:

$Pr(alpha|beta, gamma) = frac{Pr(alpha) Pr(beta|alpha) Pr(gamma|alpha)}{Pr(beta, gamma)}$

Three probabilities in the numerator are given but the denominator is not, then the given constraints are not sufficient to calculate $Pr(alpha|beta, gamma)$.

Am I right?

probability conditional-probability

probability conditional-probability

share|cite|improve this question

asked Jan 17 at 14:35

lernerlerner

314217

share|cite|improve this question

asked Jan 17 at 14:35

lernerlerner

314217

share|cite|improve this question

share|cite|improve this question

asked Jan 17 at 14:35

lernerlerner

314217

asked Jan 17 at 14:35

lernerlerner

314217

asked Jan 17 at 14:35

lernerlerner

314217

314217

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

1

$begingroup$

You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$

P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.

share|cite|improve this answer

edited Jan 17 at 14:50

answered Jan 17 at 14:47

HarnakHarnak

1,299512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
  $endgroup$
  – lerner
  Jan 17 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need either tables or an analytical expression in function of $alpha$.
  $endgroup$
  – Harnak
  Jan 17 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries. I was about to ask what you meant :)
  $endgroup$
  – Harnak
  Jan 18 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
  $endgroup$
  – lerner
  Jan 19 at 3:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
  $endgroup$
  – Harnak
  Jan 21 at 20:46
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

Thanks for contributing an answer to Mathematics Stack Exchange!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077053%2fare-these-numbers-sufficient-for-calculating-p-alpha-beta-gamma%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$

P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.

share|cite|improve this answer

edited Jan 17 at 14:50

answered Jan 17 at 14:47

HarnakHarnak

1,299512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
  $endgroup$
  – lerner
  Jan 17 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need either tables or an analytical expression in function of $alpha$.
  $endgroup$
  – Harnak
  Jan 17 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries. I was about to ask what you meant :)
  $endgroup$
  – Harnak
  Jan 18 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
  $endgroup$
  – lerner
  Jan 19 at 3:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
  $endgroup$
  – Harnak
  Jan 21 at 20:46
  

  
  
  
  

add a comment | 

1

$begingroup$

You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$

P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.

share|cite|improve this answer

edited Jan 17 at 14:50

answered Jan 17 at 14:47

HarnakHarnak

1,299512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
  $endgroup$
  – lerner
  Jan 17 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need either tables or an analytical expression in function of $alpha$.
  $endgroup$
  – Harnak
  Jan 17 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries. I was about to ask what you meant :)
  $endgroup$
  – Harnak
  Jan 18 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
  $endgroup$
  – lerner
  Jan 19 at 3:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
  $endgroup$
  – Harnak
  Jan 21 at 20:46
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$

P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.

share|cite|improve this answer

edited Jan 17 at 14:50

answered Jan 17 at 14:47

HarnakHarnak

1,299512

$endgroup$

You could use Bayes Theorem and then use the independence given $alpha$:
$$mathbb{P}(beta, gamma) = sum_{alpha} mathbb{P}(beta, gamma | alpha)mathbb{P}(alpha)= sum_{alpha}mathbb{P}(beta | alpha) mathbb{P} (gamma | alpha) mathbb{P} (alpha) $$

P.S.: I'm assuming you are given those probabilities for each $alpha$ and $alpha$ spans in a discrete set.

share|cite|improve this answer

edited Jan 17 at 14:50

answered Jan 17 at 14:47

HarnakHarnak

1,299512

share|cite|improve this answer

share|cite|improve this answer

edited Jan 17 at 14:50

edited Jan 17 at 14:50

edited Jan 17 at 14:50

answered Jan 17 at 14:47

HarnakHarnak

1,299512

answered Jan 17 at 14:47

HarnakHarnak

1,299512

answered Jan 17 at 14:47

HarnakHarnak

1,299512

1,299512

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
  $endgroup$
  – lerner
  Jan 17 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need either tables or an analytical expression in function of $alpha$.
  $endgroup$
  – Harnak
  Jan 17 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries. I was about to ask what you meant :)
  $endgroup$
  – Harnak
  Jan 18 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
  $endgroup$
  – lerner
  Jan 19 at 3:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
  $endgroup$
  – Harnak
  Jan 21 at 20:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
  $endgroup$
  – lerner
  Jan 17 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, you need either tables or an analytical expression in function of $alpha$.
  $endgroup$
  – Harnak
  Jan 17 at 14:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No worries. I was about to ask what you meant :)
  $endgroup$
  – Harnak
  Jan 18 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
  $endgroup$
  – lerner
  Jan 19 at 3:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
  $endgroup$
  – Harnak
  Jan 21 at 20:46
  

  
  
  
  

$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57

$begingroup$
I think the Bayesian tables are given, then $P(beta,gamma)$ becomes calculable.
$endgroup$
– lerner
Jan 17 at 14:57

$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59

$begingroup$
Yes, you need either tables or an analytical expression in function of $alpha$.
$endgroup$
– Harnak
Jan 17 at 14:59

1

1

$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19

$begingroup$
No worries. I was about to ask what you meant :)
$endgroup$
– Harnak
Jan 18 at 15:19

$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32

$begingroup$
Is it right that we can know the probability for each $alpha$ if $Pr(alpha)$ is given as described in the question?
$endgroup$
– lerner
Jan 19 at 3:32

$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46

$begingroup$
Actually, I assumed you were given $mathbb{P}(alpha)$ for every $alpha$. Otherwise you cannot apply Bayes Theorem.
$endgroup$
– Harnak
Jan 21 at 20:46

add a comment | 

Thanks for contributing an answer to Mathematics Stack Exchange!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Thanks for contributing an answer to Mathematics Stack Exchange!

• Please be sure to answer the question. Provide details and share your research!

But avoid …

• Asking for help, clarification, or responding to other answers.


• Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077053%2fare-these-numbers-sufficient-for-calculating-p-alpha-beta-gamma%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

This page is only for reference, If you need detailed information, please check here